Added a few generic problems\nSolved Euler027

This commit is contained in:
FuryFire 2016-02-23 15:55:26 +01:00
parent a6355b2629
commit 2a2623a652
19 changed files with 352 additions and 0 deletions

@ -0,0 +1,53 @@
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BIN
Generic/Fibonacci/a.exe Normal file

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Generic/Fibonacci/solve.c Normal file

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#include <stdio.h>
#include <string.h>
#define SIZE 20
void add(char *a, char *b, char *result, int size) {
int i = 0;
for(i=0;i<size;i++) {
//We have overflow!
if((a[i] + b[i] + result[i] < a[i]) || (a[i] + b[i] + result[i] < b[i])) {
result[i+1]++;
}
result[i]+= a[i] + b[i];
}
}
int main( )
{
unsigned char prev[SIZE];
unsigned char current[SIZE];
unsigned char next[SIZE];
unsigned int i=0;
memset(prev,0x00,SIZE);
memset(current,0x00,SIZE);
memset(next,0x00,SIZE);
prev[0]=1;
current[0]=1;
next[0]=1;
for(i=2;i<100;i++) {
add(current,prev,next,SIZE);
memcpy(prev,current,SIZE);
memcpy(current,next,SIZE);
//prev = current;
//current = next;
int j;
for(j=0;j<SIZE;j++) {
printf("%d: %i\n",j,current[j]);
}
}
for(i=0;i<SIZE;i++) {
printf("%d: %i\n",i,current[i]);
}
return 0;
}

BIN
Generic/Fibonacci/solve.exe Normal file

Binary file not shown.

@ -0,0 +1 @@
console.log("Hello World");

@ -0,0 +1,20 @@
#include <stdio.h>
#define SIZE_X 12
#define SIZE_Y 12
void main()
{
int x,y;
int MemorizeThis[SIZE_X][SIZE_Y];
for (x =1; x <= SIZE_X; x++)
{
for (y = 1; y <= SIZE_Y; y++)
{
MemorizeThis[x-1][y-1] = x * y;
printf("[%03d]", MemorizeThis[x-1][y-1]);
}
printf("\n");
}
printf("\n");
return;
}

10
Generic/Seive/desc.yml Normal file

@ -0,0 +1,10 @@
title: Sieve of Eratosthenes
url: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
desc:Generate primes from 2 to 100
solution: Do i need to say more
solutions:
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c

22
Generic/Seive/solve.c Normal file

@ -0,0 +1,22 @@
#include <stdio.h>
void main( )
{
unsigned char primes[100];
int p;
int j;
memset(&primes, 1, 50 );
for( p = 2; p<=10; p++) {
if (primes[p]) {
for(j = p*p; j <= 100; j+= p){
primes[j] = 0;
}
}
}
printf( "Primes:%d\n", sizeof(primes)) ;
for(p=0;p<sizeof(primes);p++) {
if(primes[p]) {
printf( "%d\n",p);
}
}
}

23
ProjectEuler/026/desc.yml Normal file

@ -0,0 +1,23 @@
title: Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
url: http://projecteuler.net/problem=26
desc: |
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
solution: Bruteforce
solutions:
solve.php:
desc: Basic Solution - needs BCMath
language: php

@ -0,0 +1,20 @@
<?php
define('MAX',1000);
$max_len = 0;
for($div = MAX; $div > 2; $div--) {
$result = bcdiv('1',(string)$div,2000);
for($len = 2; $len < 1000; $len++) {
if(substr($result,10,$len) === substr($result,10+$len,$len)) {
if($len > $max_len) {
echo "Div: ".$div. " Len: ".$len."\n";
$max_len = $len;
$max_div = $div;
}
break;
}
}
}
echo $max_div;

13
ProjectEuler/026/solve.rb Normal file

@ -0,0 +1,13 @@
DIGITS = 1000
max = 10 ** (DIGITS-1)
fcurrent = 1
fprev = 1
term = 2
while fcurrent < max do
term = term + 1
fnext = fcurrent + fprev
fprev = fcurrent;
fcurrent = fnext;
end
puts term

20
ProjectEuler/027/desc.yml Normal file

@ -0,0 +1,20 @@
title: Quadratic primes
url: http://projecteuler.net/problem=27
desc: |
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0
solution: Bruteforce
solutions:
solve.php:
desc: Basic Solution - needs BCMath
language: php

@ -0,0 +1,32 @@
<?php
function is_prime($prime) {
if($prime == 1)
return true;
if($prime == 2)
return true;
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
$high = 0;
$max_primes = 0;
for($a=-999; $a<1000; $a++) {
for($b=-999; $b<1000; $b=$b+2) {
$n=0;
while(true) {
echo $n*$n + $a * $n + $b . "\n";
if(!is_prime($n*$n + $a * $n + $b)) {
if($n > $max_primes) {
$max_primes = $n;
echo "max: $n a: $a b: $b\n";
}
break;
}
$n++;
}
}
}

13
ProjectEuler/027/solve.rb Normal file

@ -0,0 +1,13 @@
DIGITS = 1000
max = 10 ** (DIGITS-1)
fcurrent = 1
fprev = 1
term = 2
while fcurrent < max do
term = term + 1
fnext = fcurrent + fprev
fprev = fcurrent;
fcurrent = fnext;
end
puts term

@ -0,0 +1,5 @@
copy.src.files=false
copy.src.target=
index.file=
run.as=SCRIPT
url=http://localhost/codingtests/

@ -0,0 +1,7 @@
include.path=${php.global.include.path}
php.version=PHP_54
source.encoding=UTF-8
src.dir=.
tags.asp=false
tags.short=true
web.root=.

9
nbproject/project.xml Normal file

@ -0,0 +1,9 @@
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://www.netbeans.org/ns/project/1">
<type>org.netbeans.modules.php.project</type>
<configuration>
<data xmlns="http://www.netbeans.org/ns/php-project/1">
<name>codingtests</name>
</data>
</configuration>
</project>